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Infinitesimal displacement in spherical coordinates

  1. May 25, 2015 #1
    I'm trying to derive what ##ds^2## equals to in spherical coordinates.
    In Euclidean space, $$ds^2= dx^2+dy^2+dz^2$$
    Where ##x=r \ cos\theta \ sin\phi## , ##y=r \ sin\theta \ sin\phi## , ##z=r \ cos\phi## (I'm using ##\phi## for the polar angle)
    For simplicity, let ##cos \phi = A## and ##sin \phi = B##
    Then ##dx=r \ cos\theta \ dB + B(cos \theta \ dr - r \ sin \theta \ d \theta)##
    ##dy=r \ sin\theta \ dB + B(sin \theta \ dr + r \ cos \theta \ d \theta)##
    ##dz= r \ dA + A \ dr##
    When I add up the squares of the 3 differential elements above, gather similar terms and replace ##A## and ##B##, I get ##ds^2 = dr^2 + r^2 \ sin^2 \phi \ d \theta^2## , and you can see that one term is missing. Is something wrong with my approach?
     
  2. jcsd
  3. May 25, 2015 #2
    I'm using ##\phi ## as the polar angle and ##\theta## as the azimuthal angle over here. Either way, that won't add the missing term, but will only interchange the angle variables.
     
  4. May 25, 2015 #3
    Sorry I see later.
    $$ x = rB\cos{theta} \,\Rightarrow\, dx = dr\,B\cos{\theta} + r\,dB\,\cos{theta} - rB\sin{\theta} = (B\,dr + r\,dB)\cos{\theta} - rB\sin{\theta} $$
    $$ dy = (B\,dr + r\,dB)\sin{\theta} +rB\cos{\theta} \,\text{and}\, dz = A\,dr-rB\,d\phi $$
    because
    $$ A = cos{\phi} \Rightarrow dA = -\sin{\phi}\,d\phi = -B\,d\phi \,\text{and}\, dB = A\,d\phi $$
    I think your problem is that A depends by B.
     
    Last edited: May 25, 2015
  5. May 28, 2015 #4
    I found an error in my steps. I'm getting the correct answer now. Anyways, thanks for helping :)
     
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